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noname [10]
3 years ago
13

P6+8=11 what is the value of p

Mathematics
2 answers:
scoundrel [369]3 years ago
8 0

Answer:

p=\frac{1}{2}

Step-by-step explanation:

p6+8=11\\\\p6+8-8=11-8\\\\p6=3\\\\\frac{p6=3}{6}\\\\\boxed{p=\frac{1}{2}}

Hope this helps.

Vaselesa [24]3 years ago
5 0

Answer:

p = 1/2

Step-by-step explanation:

6p + 8 = 11

6p = 11 - 8

6p = 3

6p/6 = 3/6

p = 1/2

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$256

Step-by-step explanation:

Each class got $256. I know this because I did 3,840 divided by 15 which got me the CORRECT answer of $256

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A survey asks, "If the husband in a family wants children, but the wife decides that she does not want any children, is it all r
vodka [1.7K]

Answer:

a. The 99% confidence interval for the population proportion is (0.7872, 0.8610).

D. Yes, we can conclude that the population proportion exceeds 75% because 75% is below the lowest believable value of the confidence interval.

b. The 99% confidence interval would be wider than a 95% confidence interval.

As the confidence level increases, the width interval increases, as we are requiring more confidence with the same information (there is no new sample). This means that, to be more confident, the only way is to include more values in the interval.

Step-by-step explanation:

We have to calculate a 99% confidence interval for the proportion.

The sample proportion is p=0.8241.

p=X/n=581/705=0.8241

The standard error of the proportion is:

\sigma_p=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.8241*0.1759}{705}}\\\\\\ \sigma_p=\sqrt{0.000206}=0.0143

The critical z-value for a 99% confidence interval is z=2.5758.

The margin of error (MOE) can be calculated as:

MOE=z\cdot \sigma_p=2.5758 \cdot 0.0143=0.0369

Then, the lower and upper bounds of the confidence interval are:

LL=p-z \cdot \sigma_p = 0.8241-0.0369=0.7872\\\\UL=p+z \cdot \sigma_p = 0.8241+0.0369=0.8610

The 99% confidence interval for the population proportion is (0.7872, 0.8610).

We can conclude that there is, at least, 99% chances that the true proportion is higher than 0.7872. So there is at least 99% chances that the population proportion is higher than 0.75.

4 0
3 years ago
You have $480 in your bank account.
Orlov [11]

Answer: $203

Step-by-step explanation:

e=electric bill

480-e=277

480-277=e

e=$203

8 0
2 years ago
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A machine is shut down for repairs if a random sample of 100 items selected from the daily output of the machine reveals at leas
Harlamova29_29 [7]

Answer:

0.7995 = 79.95% probability that the sample will contain at least three defectives.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

Suppose that a random sample of 20 items is selected from the machine.

This means that n = 20

The machine produces 20% defectives

This means that p = 0.2

Mean and standard deviation:

\mu = E(X) = np = 20*0.2 = 4

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.2*0.8} = 1.79

Probability that the sample will contain at least three defectives

Using continuity correction, this is P(X \geq 3 - 0.5) = P(X \geq 2.5), which is 1 subtracted by the pvalue of Z when X = 2.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{2.5 - 4}{1.79}

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Z = -0.84 has a pvalue of 0.2005

1 - 0.2005 = 0.7995

0.7995 = 79.95% probability that the sample will contain at least three defectives.

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3 years ago
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